From the figure,
sin \(\theta\) = \(\frac { AB }{ AC } \)
sin 48⁰30' = \(\frac { AB }{ 5 } \)
0.7490 = \(\frac { AB }{ 5 } \)
5 \(\times\) 0.7490 = AB
AB = 3.7450 cm
cos \(\theta\) = \(\frac { BC }{ AC } \)
cos 48⁰30' = \(\frac { BC }{ 5 } \)
0.6626 = \(\frac { BC }{ 5 } \)
0.6626 \(\times\) 5 = BC
BC  = 3.313 cm
Area of right triangle = \(\frac { 1 }{ 2 } \) bh
\( =\frac{1}{2} \times B C \times A B \)
\( =\frac{1}{2} \times 3.3130 \times 3.7450 \)
\( =1.6565 \times 3.7450 \) = 6.2035925 cm²

(i) We know that cosec A =  sec(90⁰ A)
sec(90⁰ - A) sec(34⁰)
90⁰ - A =34⁰
We get A = 90° − 34°
A = 56⁰
(ii) We know that tan B = cot(90⁰ - B)
cot(90⁰ - B) = cot 47⁰
90⁰ - B = 47⁰
We get B = 90° − 47°
B = 43⁰

(i) \(\frac { sin\ 49° }{ cos\ 41° } \) 
sin 49⁰ = sin(90⁰ - 41⁰) = cos 41⁰,  since 49⁰ + 41⁰ = 90⁰ (complementary),
Hence on substituting sin 49^o = cos41^o we get, \( \frac { cos\ 41° }{ cos\ 41° } \)= 1
(ii) \(\frac { sec\ 63° }{ cosec\ 27° } \)
sec63^o = sec (90^o- 27^o) = cosec27^o, here, 63^o and 27^o are complementary angles
we have \(\frac { sec\ 63° }{ cosec\ 27° } =\frac { cosec\ 27° }{ cosec\ 27° } =1\)

In the given right angled triangle, note that for the given angle \(\theta \), PR is the ‘opposite’ side and PQ is the ‘adjacent’ side.
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { PR }{ QR } =\frac { 35 }{ 37 } \)
\(cos\theta =\frac { adjacent\ side }{ hypotenuse } =\frac { PQ }{ QR } =\frac { 12 }{ 37 } \)
\(tan\theta =\frac { opposite\ side }{ adjacent\ side } =\frac { PR }{ PQ } =\frac { 35 }{ 12 } \)
It is enough to leave the ratios as fractions. In case, if you want to simplify each ratio neatly in a terminating decimal form, you may opt for it, but that is not obligatory.

Hypotenuse = \(\sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } \)
= \(\sqrt { 144+25\quad } =\sqrt { 69 } \)
= 13
\(sin\theta =\cfrac { 5 }{ 13 } ;cos\theta =\cfrac { 12 }{ 13 } ;tan\theta =\cfrac { 5 }{ 12 } ; \)
\(cosec\theta =\cfrac { 13 }{ 5 } ;sec\theta =\cfrac { 13 }{ 12 } ;cot\theta =\cfrac { 12 }{ 3 } \)

\(cot\ A=\cfrac { 2 }{ 3 } \)
\(cot\ A=\cfrac { Adjacent\ side }{ Opposite\ side } \)
\(tan\ A=\ \cfrac { Opp.side }{ Adj.side } =\cfrac { 3 }{ 2 } \)
\(sin\ A=\cfrac { 3 }{ \sqrt { 13 } } \)
\(cotA=\cfrac { 2 }{ \sqrt { 13 } } \)
\(\therefore \cfrac { 4sinA-3cosA }{ 2ainA+3cosA } =\cfrac { 4\times \cfrac { 3 }{ \sqrt { 13 } } -3\times \cfrac { 2 }{ \sqrt { 13 } } }{ 2\times \cfrac { 3 }{ \sqrt { 13 } } +3\cfrac { 2 }{ \sqrt { 13 } } } \)
= \(\cfrac { \frac { 12 }{ \sqrt { 13 } } -\frac { 6 }{ \sqrt { 13 } } }{ \frac { 6 }{ \sqrt { 13 } } +\frac { 6 }{ \sqrt { 13 } } } =\cfrac { \frac { 6 }{ \sqrt { 13 } } }{ \frac { 2 }{ \sqrt { 13 } } } =\cfrac { 6 }{ \sqrt { 13 } } \times \cfrac { \sqrt { 13 } }{ 12 } \)
= \(\cfrac { 6 }{ 12 } =\cfrac { 1 }{ 2 } \)

By Pythagoras theorem In\(\triangle\)OAB,
AB² = OA²+ OB²
= 3² + 2²
= 9+4
= 13
\(AB=\sqrt { 13 } \)

\(sin\ \theta =\cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(cos\ \theta =\cfrac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(b\ sin\ \theta =b\times \cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } -\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ... (1)
\(a\ cos\ \theta =a\times \cfrac { b }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } } =\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ...(2)
(1) = (2)\(\Rightarrow\) LHS = RHS
Hence proved

By By the Pythagoras theorem
BC² = AC² + AB²
= \(\left( \sqrt { { a }^{ 2 }+{ b }^{ 2 } } \right) ^{ 2 }-{ a }^{ 2 }\)
= a²+b²-a²
= b²
BC = b

By the pythagoras theorem,
AB² = OA² + OB²
(1 + x²)² = (2x)² + OB²
OB² = (1 +x²)² - (2x)² = 1+ x⁴ + 2x² - 4x² = 1+x⁴ - 2x²
OB² = (1-x²)² B
OB = (1-x²)
\(\therefore sin\ A=\cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(tan\ A=\cfrac { 1-{ x }^{ 2 } }{ 2x } \)

(i) (cos0⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ - cos 45⁰ + cos 60⁰)
= \(\left[ 1+\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \left[ 1-\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \)
= \(\left[ \frac { 2\sqrt { 2 } +2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] \left[ \frac { 2\sqrt { 2 } -2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] =\left[ \frac { 3\sqrt { 2 } +2 }{ 2\sqrt { 2 } } \right] \left[ \frac { 3\sqrt { 2 } -2 }{ 2\sqrt { 2 } } \right] \)
= \(\frac { 18-4 }{ 4\left( \sqrt { 2 } \right) ^{ 2 } } =\frac { 14 }{ 4\times 2 } =\frac { 7 }{ 4 } \) 
(ii) tan²60⁰ - 2tan²45⁰ - cot²30⁰ +2sin²30⁰ + \(\frac { 3 }{ 4 } \) cosec²45⁰
= \(\left( \sqrt { 3 } \right) ^{ 2 }-2(1)^{ 2 }-\left( \sqrt { 3 } \right) ^{ 2 }+2\left( \frac { 1 }{ 2 } \right) ^{ 2 }+\frac { 3 }{ 4 } \left( \sqrt { 2 } \right) ^{ 2 }\)
= \(3-2-3+\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \)
= -2 + \(\frac { 4 }{ 2 } \) = -2 + 2 = 0